TI-84 Plus CE: Beam Pattern for Uniform Array

TI-84 Plus CE: Beam Pattern for Uniform Array

Introduction

The following formula calculates an antenna’s normalized beam response in decibels given the following:

* the number of sensors of an antenna

* the wavelength of the beam pattern

* the angle incident of the wavelength from the ground, in degrees.

* the steering angle of the array, in degrees; which is measured from the zenith, starting directly upwards with the angle going downwards.

The program calculates the response in from a range of incident angles.

R(ΘI) = [ sin(180 * n * d * (sin(ΘI) – sin(ΘS) ÷ λ ] ÷ [ n * sin(180 * d * (sin(ΘI) – sin(ΘS) ÷ λ ]

If the denominator is 0, then the response is 0 dB.

Variables:

n = number of sensors

d = spacing between sensors

λ = wavelength

ΘI = incident angle

ΘS = steering angle

TI-84 Plus CE Code: BEAMPTN

ClrHome

Degree

Disp “BEAM PATTERN”, “BY STEPHEN A HERTZ”, “(HP 67 ANTENNAS)”

Input “NO. SENSORS? “, N

Input “INTERIOR SPACING? “, D

Input “WAVELENGTH? “, L

Input “STEERING Θ°? “, S

Input “STARTING ΘI? “, I

Input “NO. PTS? “, K

Input “CHG Θ°? “, J

For(Θ, I, I+K*J, J)

180 * D / L → A

sin(Θ) – sin(S) → B

N * sin(A * B) → R

If R=0

0 → R

Else

sin(N * A * B) / R → R

20 * log(abs(R)) → R

End

ClrHome

Disp “ΘI=”, Θ, “R (DB)=”, R

Pause

End

Example

Input:

n = 6

d = 36 ft

λ = 100 ft

ΘS = 0° (steering angle)

Range: ΘI = starting at 0°, 5 calculations, in increments of 5°

Results:

K =	ΘI =	R(ΘI) =
0	0	0
1	5	-0.4983317822
2	10	-2.056358346
3	15	-4.9617921711
4	20	-9.809670689

The negative response signifies that there is a reduction of signal strength.

Source

Hertz, Stephen A. “Beam Pattern for Uniform Array” HP-67/HP-97: Users’ Library Solutions: Antennas. Hewlett Packard. Corvallis, OR. Rev. D. April 1979. pp. 24-27

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

TI-84 Plus CE: Signal to Noise Ratio and Shannon’s Law

TI-84 Plus CE: Signal to Noise Ratio and Shannon’s Law

Signal to Noise Ratio (SNR)

The program SNR calculates the signal to noise ratio for inputs in one of the three units:

Decibels	S – N
Watts	20 * log( S ÷ N )
Voltage	10 * log(S ÷ N )

S: desired signal level

N: noise level

Cadence PCB Solutions (see source) provides a scale of signal to noise ratios and their effectiveness, which is summarized here:

5 dB to 10 dB	No connection is made, the ratio is too low
10 dB to 15 dB	The connection is unreliable
15 dB to 25 dB	Minimum acceptable level
25 dB to 40 dB	Good connection
41 dB and above	Excellent connection

In general, the higher the SNR, the better.

TI-84 TI-Basic Code: Signal to Noise Ratio

Program Title: SNR

Menu(“SIGNAL NOISE RATIO”,”DECIBELS”,1,”WATTS”,2,”VOLTAGE”,3)

Lbl 1

Input “SIGNAL DB? “, S

Input “NOISE DB? “, N

S – N → R

Goto 0

Lbl 2

Input “SIGNAL (W)? “, S

Input “NOISE (W)? “, N

20 * log(S / N) → R

Goto 0

Lbl 3

Input “SIGNAL (V)? “, S

Input “NOISE (W)? “, W

10 * log(S / N) → R

Goto 0

Lbl 0

Disp “SNR (DB): “, R

Examples

Example 1: S: -10 dB, N: -50 dB (from Source)

Results: 40 dB

Example 2: S: 400 W, N: 60 W

Results: 16.47817482 dB

Example 3: S: 300E-3 V (300 millivolts), N: 2E-6 V (2 microvolts)

Result: 51.76091259 dB (*source erroneously had 62 DB)

Shannon’s Law

The program SHANNON makes two calculations: signal to noise ratio in decibels and the capacity of the channel in bits per second. Shannon’s Law was discovered by Claude Shannon during World War II.

C = W * log(1 + S ÷ N) ÷ log(2)

W: bandwidth of the signal in Hertz

S: average signal received in Watts

N: average noise signal in Watts

C: maximum channel capacity in Bits per Second

TI-84 TI-Basic Code: Shannon’s Law

Program Title: SHANNON

Disp “SHANNON’S LAW”

Input “REC’D POWER (W)? “, S

Input “NOISE POWER (W)? “, N

Input “BANDWIDTH (HZ)? “, W

20 * log(S / N) → R

Disp “SIGNAL NOISE RATIO: “, R

W * log(1 + S / N) / log(2) → C

Disp “CHANNEL (BITS/S):”, C

Example

Inputs:

Average Received Signal Power (S): 49.2 W

Average Noise Power (N): 2.2 W

Bandwidth: 200 Hz

Results:

Signal Noise Ratio (SNR): 26.99084844 dB

Channel capacity: 909.2385861 bits/s

Source

Cadence. “What is Signal to Noise Ratio and How to calculate it?” Cadence PCB Solutions. 2020. Accessed November 30, 2025. https://resources.pcb.cadence.com/blog/2020-what-is-signal-to-noise-ratio-and-how-to-calculate-it

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

TI-84 Plus CE and Swiss Micros DM32: Blood Alcohol Level

TI-84 Plus CE and Swiss Micros DM32: Blood Alcohol Level

Just in time for the upcoming St. Patrick’s Day in a few weeks…

Introduction

The blood alcohol level is calculated by the following approximation by Walter L. Gregory Jr. using data from the CRC Handbook of Tables for Applied Engineering (1970). The program is based off the HP-97/HP-67 program listed in their Medical Practitioner pac (1979). (see source):

BAC ≈ ((ALC * OZ) ÷ 50 – T) * (3.751 ÷ WT)

ALC: Percentage of alcohol consumed. Enter as a whole number. For example, if the drink has 40% alcohol, ALC = 40. If the alcohol is measured in proof, divide the proof by 2.

OZ: The total amount of ounces consumed within the period of time. For example: six 12-oz beers total 72 ounces.

WT: The weight of the person in pounds.

T: The number of hours drinks are consumed. If T > 1, then T = hours – 1. Otherwise T = 0.

If calculated BAC is below 0, then set BAC = 0.

TI-84 Program CE: ALCOHOL

Type: TI Basic

Disp “BLOOD ALCOHOL (US)”

Input “WEIGHT (LB)? “, W

Input “OUNCES? “, O

Input “PERCENTAGE? “, P

Input “HOURS OF DRINKING? “, H

If H≤1 : Then : 0 → H

Else : H – 1 → H : End

(P * O / 50 – H) * 3.751 / W → A

Disp “BLOOD ALCOHOL: “, A

Swiss Micros DM32 (HP 32SII): ALCOHOL

A01 LBL A

A02 INPUT W

A03 INPUT O

A04 INPUT P

A05 INPUT H

A06 RCL H

A07 1

A08 x≥y?

A09 SF 1

A10 STO- H

A11 FS? 1

A12 Cl x

A13 FS? 1

A14 STO H

A15 CF 1

A16 RCL P

A17 RCL× O

A18 50

A19 ÷

A20 RCL- H

A21 3.751

A22 RCL÷ W

A23 ×

A24 x<0?

A25 Cl x

A26 RTN

Notes:

A07 1, A08 x≥y?, A09 SF 1: If hours > 1, set Flag 1. Using flags can sometimes eliminates the need for additional labels.

A24 x<0?, A25 Cl x: If the number in the display (x stack) negative, change it to zero.

Examples

Example 1:

Inputs:

Weight: W = 150 lbs

Ounces: O = 4 oz

Percentage: P = 20%

Hours: H = 0.5 hours (T = 0)

Output: 0.0400

Example 2:

Inputs:

Weight: W = 180 lbs

Ounces: O = 45 oz

Percentage: P = 50%

Hours: H = 2.5 hours (T = 1.5)

Output: 0.9065

(typo:  Thank you, Pedro Leiva)

Be sure to check your local laws for legal blood alcohol levels. This program is not suited to be evidence in legal matters.

One thing remains: DON’T DRINK AND DRIVE

Source

Hewlett Packard. HP-67/HP-97. User’s Library Solutions: Medical Practitioner. Corvallis, OR. April 1979. pp. 9-12

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

TI-84 Plus CE: Logistic Map

TI-84 Plus CE: Logistic Map

It feels like forever since I with the TI-84 Plus CE.

Introduction

The program LOGISTIC plots the sequence

x = r * x * (1 – x)

where x is initially in the interval (0, 1) and r is a parameter. The sequence traces out the path of recursion.

Depending on the value of r, the sequence can eventually stabilize to a single point, fluctuate then stabilize, eventually bounce between two stable points, or going into the chaos. The higher the r, chance of the latter two possibilities increase. Typically, for any r < 3, the sequence will most likely stabilize.

TI-84 Plus CE Program: LOGISTIC.8xp

Type: TI-Basic Program

ClrHome

Disp “LOGISTIC MAP”

Disp “X1=R*X0*(1-X0)”

Seq

FnOff

Menu(“LAMBDA (R)”, “ENTER”, 1, “RANDOM”, 2)

Lbl 1

Input “0<R≤4: “, R

Goto A

Lbl 2

randInt(1,80) / 20 → R

Pause R

Goto A

Lbl A

Menu(“U(1)=”, ”ENTER”, 3, “RANDOM”, 4)

Lbl 3

Input “U(1)? “, U

Goto B

Lbl 4

rand → U

Pause U

Goto B

Lbl B

{U} → u(nMin)

SEQ(n+1)

“R*u(n)*(1-u(n))” → u

FnOn 1

1 → nMin

50 → nMax

0 → Ymin

1 → Ymax

0 → Xmin

50 → Xmax

DispGraph

Notes:

The Time setting is assumed: x-axis = n, y-axis = u__n

nMin: Sets the counter and the index of the first argument. Typically the counter starts with 0 or 1. Keystrokes: [ vars ], 1. Window, scroll to U/V/W, 4. nMin.

Seq: Sets Sequence mode.

I use R for λ since the TI-84 does not have Greek characters (except for π and σ).

randInt(1,80) / 20 → R: Selects random values from 0.05 and 4 with a tick mark of 0.05.

u(nMin): Store the initial conditions. The conditions are entered as a list, even there is only one initial condition. Keystrokes: [ vars ], 1. Window, scroll to U/V/W, 1. u(nMin)

SEQ(n+1): In the program editor, I had to select this from the catalog. It is listed under “SEQ(n+1) Type”. This allows for the recurring sequence to be in the format u(n+1) = f(u(n),n).

“R*u(n)*(1-u(n))” → u: The u is type from the keyboard: [ 2^nd ] [ 7 ].

FnOn 1: Turns the sequence u(n) on.

Examples

Left: r = 3.67, initial point = 0.2. This looks like total chaos as n grows.

Right: r = 3.1, initial point = 0.5. This sequence does not converge to a single value, but eventually provides a bifurcation between two points.

Caution: If r is too high, even if the u(1) is with in the interval [0, 1], the sequence can “escape” and an overflow can occur. I don’t recommend an r higher than 4.

Source

“Logistic map”. Wikipedia https://en.wikipedia.org/wiki/Logistic_map Retrieved October 20, 2025.

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

HP 21S and TI-84 Plus CE: Normal Distributions and Hypothesis Tests

HP 21S and TI-84 Plus CE: Normal Distributions and Hypothesis Tests

Let’s compare how calculations involving the normal distribution are done between a classic calculator and a current one. The classic calculator is the rare HP 21S calculator from 1988 and the current calculator is the TI-84 Plus CE.

In 2017 (how time flies!) I wrote a review for the HP 21S (along with it’s cousin HP 20S):

https://edspi31415.blogspot.com/2017/04/retro-review-hewlett-packard-hp-20s-and.html.

Normal Distribution Calculations

z = z-score, point; p = probability, area

Assumptions: The mean is assumed to be 0 while the standard deviation is 1. (μ = 0, σ = 1).

Lower Tail (-∞, z)

	HP 21S	TI-84 Plus CE*	Example
Given z, find p	Keystrokes: 1 [ - ] z [ ←| ] (Q(z)) [ = ]	normalcdf(-1E99, z)	Input: z = 0.77 Result: p ≈ 0.7794
Given p, find z	Keystrokes: [ ( ] 1 [ - ] p [ ) ] [ |→ ] (zp)	invNorm(p)	Input: p = 0.77 Result: z ≈ 0.7388

* also includes TI-83 Plus, TI-84 Plus, TI-83 Premium (Python), TI-82 Advanced, and I do believe it is the same syntax for TI-89/Nspire family

Upper Tail (z, +∞)

	HP 21S	TI-84 Plus CE*	Example
Given z, find p	Keystrokes: z [ ←| ] (Q(z))	normalcdf(z, 1E99)	Input: z = 0.49 Result: p ≈ 0.3121
Given p, find z	Keystrokes: p [ |→ ] (zp)	InvNorm(1 - p)	Input: p = 0.49 Result: z ≈ 0.0251

Two Tail (-z, z)

	HP 21S**	TI-84 Plus CE*	Example
Given z, find p	Keystrokes: z [ STO ] [ 0 ] [ +/- ] [ ←| ] (Q(z)) [ - ] [ RCL ] 0 [ ←| ] (Q(z)) [ = ]	normalcdf(-z, z)	Input: z = 1 Result: p ≈ 0.6827
Given p, find z	Keystrokes: p [ ÷ ] 2 [ = ] [ STO ] 0 Lower Limit: [ ( ] 0.5 [ - ] [ RCL ] 0 [ ) ] [ |→ ] (zp) Upper Limit: [ ( ] 0.5 [ + ] [ RCL ] 0 [ ) ] [ |→ ] (zp)	InvNorm(p, 0, 1, CENTER)	Input: p = 0.25 Result: Upper: z ≈ 0.3186 Lower: z ≈ -0.3186

** HP 21S: A memory register is needed for this particular algorithm. I use register 0 for this example.

Sample Test of the Mean: Is the Proposed Mean the True Mean? (μ0 = μ)

Perform a significant test of whether a proposed mean (μ0) is the true mean (μ) given data from a sample:

x-bar: arithmetic mean of a sample

σ: population deviation of the sample

n: sample size

In this test, the null or default hypothesis is μ =μ0, while the alternate hypothesis is μ ≠ μ0.

Your confidence level and critical level, α, are complimentary. For example, if you want a 95% confidence level, your critical level, or α, is: α = 1 – 0.95 = 0.05 (5%). This is a two-tail test.

HP 21S	TI-84 Plus CE
[ ←| ] [ ← ] (LOAD) (A) (1-Stat) n [ XEQ ] D μ [ R/S ] σ [ R/S ] Test Method # 1: α [ ÷ ] 2 [ = ] [ |→ ] (zp) (= critical value) μ0 [ XEQ ] B (= test value) Null: μ0 = x-bar, Alternative: μ0 ≠ x-bar If test value < critical value, do not reject null hypothesis If test value > critical value, reject null hypothesis and accept the alternate hypothesis Test Method # 2: μ0 [ XEQ ] B [ ←| ] [ 1 ] (Q(z)) (= p) Null: μ0 = x-bar, Alternative: μ0 ≠ x-bar If p > α / 2, do not reject null hypothesis If p < α / 2, reject null hypothesis and accept the alternate hypothesis Note: LBL D: store summary statistics for the test of one sample mean or probability. For this test, the standard error that is shown will not be used. LBL B: calculate the z-score: z = (x-bar – μ0) / (σ / √n)	[ stat ], TESTS, 1: Z-Test… Inpt: Stats (Input shown as Inpt) μ0: enter μ0 σ: enter σ x-bar: enter x-bar n: enter n Select μ ≠ μ0 (alternate hypothesis) Select Calculate Test Method # 1: critical value: normalcdf(α/2, 1E99) test value: [ vars ], 5: Statistics…, TEST, 2: z Null: μ0 = μ, Alternative: μ0 ≠ μ If test value < critical value, do not reject null hypothesis If test value > critical value, reject null hypothesis and accept the alternate hypothesis Test Method # 2: Compare p to α Null: μ0 = x-bar, Alternative: μ0 ≠ x-bar If p > α, do not reject null hypothesis If p < α, reject null hypothesis and accept the alternate hypothesis One line command: Z-Test(μ0, σ, x-bar, n, alt) alt: -1 for μ0 < μ, 0 for μ0 ≠ μ, 1 for μ0 > μ
Example: n = 515, μ = 7.65, σ = 3.56 α = 5% = 0.05, μ0 = 7.7, α / 2 = 0.025 [ ←| ] [ ← ] (LOAD) (A) 515 [ XEQ ] D 7.65 [ R/S ] 3.56 [ R/S ] 0.05 [ ÷ ] 2 [ = ] [ |→ ] (zp) (critical value = 1.9600) 7.7 [ XEQ ] B (z ≈ -0.3187 < 1.9600) 7.7 [ XEQ ] B [ ←| ] [ 1 ] (Q(z)) (p ≈ 0.6250 > 0.025) Do not reject null hypothesis	Example: n = 515, μ = 7.65, σ = 3.56 α = 5% = 0.05, μ0 = 7.7 [ stat ], TESTS, 1: Z-Test… Inpt: Stats μ0: 7.7 σ: 3.56 x-bar: 7.65 n: 515 Select μ ≠ μ0 Select Calculate Results: μ≠7.7 z=-0.3187304977 p=0.7499310153 x-bar=7.65 n=515 p ≈ 0.7499 > 0.05 z ≈ -0.3187 < 1.9600 Do not reject null hypothesis

This is one of several z-tests that can be performed.

Confidence Interval

Determine a confidence level of where the true mean is located given data from a sample:

x-bar: arithmetic mean of a sample

σ: population deviation of the sample

n: sample size

Confidence Mean:

x-bar ± (z * σ) / √n

z: absolute value of the z-score where the two tail probability equals c.

Using TI-84 Plus CE commands: invNorm(c, 0, 1, CENTER)

HP 21S	TI-84 Plus CE
[ ←| ] [ ← ] (LOAD) (A) n [ XEQ ] D μ [ R/S ] σ [ R/S ] Confidential Interval: [ ( ] 1 [ - ] c [ ) ] [ ÷ ] 2 [ = ] [ |→ ] [ 1 ] (zp) [ XEQ ] C → precision [ R/S ] → lower limit [ R/S ] → upper limit Note: LBL C: Confidence interval precision = (z * σ) / √n lower limit = x-bar – precision upper limit = x-bar + precision	[ stat ], TESTS, 7: Zinterval… Inpt: Stats (Input is shown as Inpt) σ: enter σ x-bar: enter x-bar n: enter n C-Level: enter confidence level as a decimal Calculate One line command: ZInterval σ, x-bar, n, c
n = 389, μ = 88.84, σ = 10.25 c = 90% = 0.9 [ ←| ] [ ← ] (LOAD) (A) 389 [ XEQ ] D 88.84 [ R/S ] 10.25 [ R/S ] [ ( ] 1 [ - ] 0.9 [ ) ] [ ÷ ] 2 [ = ] [ |→ ] [ 1 ] (zp) [ XEQ ] C → 0.8548 [ R/S ] → 87.9852 [ R/S ] → 89.6948	n = 389, μ = 88.84, σ = 10.25 c = 90% = 0.9 Inpt: Stats σ: 10.25 x-bar: 88.84 n: 389 C-Level: 0.9 Result: (87.985, 89.695) x-bar = 88.84 n = 389

Sources

CalcBlog “Hypothesis Testing using Z-Test on the TI-83 Plus, TI-84 Plus, TI-89, and Voyage 200” February 6, 2011. https://www.calcblog.com/hypothesis-testing-z-test-on-ti84-ti89-graphing-calculator/ Accessed January 30, 2025.

Hewlett Packard HP-21 Stat/Math Calculator: Owner’s Manual Edition 3. June 1990.

TI-Basic Developer “The Z-Test( Command” http://tibasicdev.wikidot.com/z-test Accessed January 30, 2025.

Enjoy! Until next time,

Eddie

All original content copyright, © 2011-2025. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.